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The capacitance of a parallel plate arrangement is

  • A.\( \epsilon A/d \)
  • B.\( \epsilon d/A \)
  • C.\( \epsilon A d \)
  • D.\( \epsilon/(Ad) \)

Answer: A

A larger area holds more charge at a given voltage and a smaller separation gives a stronger field for the same voltage, so the capacitance rises in both cases. The formula neglects fringing at the edges, which makes the true value slightly larger.

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